Thay \(a=b=1\Rightarrow\frac{2}{8.7}\ge\frac{1}{25}\Leftrightarrow\frac{2}{56}\ge\frac{1}{25}\) (sai)

Ta có

\(\sqrt{2}\sqrt{4a+1}\le\frac{4a+3}{2}\)

\(\sqrt{2}\sqrt{4b+1}\le\frac{4b+3}{2}\)

\(\sqrt{2}\sqrt{4c+1}\le\frac{4c+3}{2}\)

\(\sqrt{2}\sqrt{4d+1}\le\frac{4d+3}{2}\)

Cộng vế theo vế ta được

\(\sqrt{2}\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1}\right)\)

\(\le8\)

<=> \(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\sqrt{4d+1}\le4\sqrt{2}\)

k mk nha

k mk nha!

#meo#

$A=\frac{64abc}{(a+b)(b+c)(c+a)}+1+\frac{16ab}{(b+c)(c+a)}+\frac{16bc}{(b+a)(c+a)}+\frac{16ac}{(a+b)(a+c)}+4.(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c})=4.(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c})+\frac{64abc}{(a+b)(b+c)(c+a)}+\frac{16ab(a+b)+16bc(b+c)+16ac(a+c)}{(a+b)(b+c)(c+a)}+1=4(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})+\frac{64abc}{(a+b)(b+c)(c+a)}+\frac{16(a+b)(b+c)(c+a)-32abc}{(a+b)(b+c)(c+a)}+1=4(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})+\frac{32abc}{(a+b)(b+c)(c+a)}+17=4\left [\frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b}+\frac{4abc}{(a+b)(b+c)(c+a)} \right ]+\frac{16abc}{(a+b)(b+c)(c+a)}+17\geq 4.2+17+\frac{16abc}{(a+b)(b+c)(c+a)}=25+\frac{16abc}{(a+b)(b+c)(c+a)}> 25$

( Do áp dụng bđt Schur mở rộng là :$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{4abc}{(a+b)(b+c)(c+a)}\geq 2$

\(\sqrt{4a+1}-2\sqrt{a}=\frac{4a+1-4a}{\sqrt{4a+1}+2\sqrt{a}}=\frac{1}{\sqrt{4a+1}+2\sqrt{a}}\)

\(\sqrt{4b+1}-2\sqrt{b}=\frac{1}{\sqrt{4b+1}+2\sqrt{b}}\)

Mà \(a>b\Rightarrow\left\{{}\begin{matrix}\sqrt{4a+1}>\sqrt{4b+1}\\2\sqrt{a}>2\sqrt{b}\end{matrix}\right.\) \(\Rightarrow\sqrt{4a+1}+2\sqrt{a}>\sqrt{4b+1}+2\sqrt{b}\)

\(\Rightarrow\frac{1}{\sqrt{4a+1}+2\sqrt{a}}< \frac{1}{\sqrt{4b+1}+2\sqrt{b}}\)

\(\Rightarrow\sqrt{4a+1}-2\sqrt{a}< \sqrt{4b+1}-2\sqrt{b}\)

a) \(\sqrt{a}+1>\sqrt{a+1}\)\(\Leftrightarrow\)\(a+2\sqrt{a}+1>a+1\)\(\Leftrightarrow\)\(2\sqrt{a}>0\)( luôn đúng \(\forall x>0\) ) 

b) \(a-1< a\)\(\Leftrightarrow\)\(\sqrt{a-1}< \sqrt{a}\)

c) \(\left(\sqrt{6}-1\right)^2=6-2\sqrt{6}+1>3-2\sqrt{3.2}+2=\left(\sqrt{3}-\sqrt{2}\right)^2\)

do \(\sqrt{6}-1>0;\sqrt{3}-\sqrt{2}>0\) nên \(\sqrt{6}-1>\sqrt{3}-\sqrt{2}\) ( đpcm ) 

Ap dung BDT Bun-hia-cop-xki ta co:

\(\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^2\le\left(1+1+1\right)\left[4\left(a+b+c\right)+3\right]=21\)

\(\Rightarrow-\sqrt{21}\le\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{21}< 5\)

\(\Rightarrow\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}< 5\)

Ap dung BDT Bun-hia-cop-xki ta co:

\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^2\le\left(1+1+1\right)\left[4\left(a+b+c\right)+3\right]=21(4a+1​+4b+1​+4c+1​)2≤(1+1+1)[4(a+b+c)+3]=21

\Rightarrow-\sqrt{21}\le\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{21}&lt; 5⇒−21​≤4a+1​+4b+1​+4c+1​≤21​<5

\Rightarrow\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}&lt; 5⇒4a+1​+4b+1​+4c+1​<5

Thôi giải lại câu 1:v (ý tưởng dồn biến là quá trâu bò! Bên AoPS em mới phát hiện ra có một cách bằng Cauchy-Schwarz quá hay!)

\(BĐT\Leftrightarrow\Sigma_{cyc}\frac{\left(a+b+c\right)^2}{2a^2+\left(a^2+b^2\right)+\left(a^2+c^2\right)}\le\frac{9}{2}\)(*)

BĐT này đúng theo Cauchy-Schwarz: \(VT_{\text{(*)}}\le\Sigma_{cyc}\left(\frac{a^2}{2a^2}+\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)=\frac{9}{2}\)

Ta có đpcm.

Equality holds when a = b = c = 1 (Đẳng thức xảy ra khi a = b =c = 1)

1/Đặt \(VT=f\left(a;b;c\right)\) và \(0< t=\frac{a+b}{2}\)

Ta có: \(f\left(a;b;c\right)-f\left(t;t;c\right)=\frac{1}{4a^2+b^2+c^2}+\frac{1}{4b^2+a^2+c^2}-\frac{2}{5t^2+c^2}+\frac{1}{a^2+b^2+4c^2}-\frac{1}{2t^2+4c^2}\)

\(=\frac{5t^2-4a^2-b^2}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)}+\frac{5t^2-4b^2-a^2}{\left(5t^2+c^2\right)\left(4b^2+a^2+c^2\right)}+\frac{2t^2-a^2-b^2}{\left(a^2+b^2+4c^2\right)\left(2t^2+4c^2\right)}\)

\(=-\frac{1}{4}\left(a-b\right)\left[\frac{\left(11a+b\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)}-\frac{\left(a+11b\right)}{\left(5t^2+c^2\right)\left(4b^2+a^2+c^2\right)}\right]+\frac{2t^2-a^2-b^2}{\left(a^2+b^2+4c^2\right)\left(2t^2+4c^2\right)}\)

Xét cái ngoặc to: \(\frac{\left(11a+b\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)}-\frac{\left(a+11b\right)}{\left(5t^2+c^2\right)\left(4b^2+a^2+c^2\right)}\)

\(=\frac{\left(11a+b\right)\left(4b^2+a^2+c^2\right)-\left(a+11b\right)\left(4a^2+b^2+c^2\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(4b^2+a^2+c^2\right)}\)

\(=\frac{\left(a-b\right)\left(7a^2-36ab+7b^2+10c^2\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(4b^2+a^2+c^2\right)}\)

Từ đó: f(a;b;c) -f(t;t;c)

\(=-\frac{\frac{1}{4}\left(a-b\right)^2\left(7a^2-36ab+7b^2+10c^2\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(4b^2+a^2+c^2\right)}+\frac{-\frac{1}{2}\left(a-b\right)^2}{\left(a^2+b^2+4c^2\right)\left(2t^2+4c^2\right)}\)

\(=-\frac{1}{4}\left(a-b\right)^2\left[\frac{\left(7a^2-36ab+7b^2+10c^2\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(4b^2+a^2+c^2\right)}+\frac{2}{\left(a^2+b^2+4c^2\right)\left(2t^2+4c^2\right)}\right]\le0\)

Do đó \(f\left(a;b;c\right)\le f\left(t;t;c\right)=f\left(t;t;3-2t\right)\)

\(=\frac{-9\left(t-1\right)^4}{2\left(3t^2-8t+6\right)\left(3t^2-4t+3\right)}+\frac{1}{2}\le\frac{1}{2}\)

Ta có đpcm.